#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
const int N = 505;
bool dp[N][500 * 100 + 5] = {false};
// dp代表从前n个数中选出的所有选法中，和为j，是否可以凑出
ll sum;
ll nums[N];
int n;
int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> nums[i];
        sum += nums[i];
    }
    if(sum%2!=0)
    {
        cout<<"false"<<endl;
    }
    else 
    {   
          dp[0][0]=true;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= sum/2; j++)
        {
            dp[i][j]=dp[i-1][j];
            if(j-nums[i-1]>=0)
            {
                dp[i][j]|=dp[i-1][j-nums[i-1]];
            }

        }
    }
    if(dp[n][sum/2]==true) cout<<"true"<<endl;
     else cout<<"false"<<endl;
    }
  
}
// 64 位输出请用 printf("%lld")